The point on the graph `y=log_2log_6{2^sqrt(2x+1)+4}` whose y coordinate is 1 is

Options

A. (1,1)
B. (6,1)
C. (8,1)
D. (12,1)

Answer

To find the point on the graph \( y = \log_2(\log_6(2^{\sqrt{2x+1}} + 4)) \) whose y-coordinate is 1, we can follow these steps:

### Step 1: Set up the equation
Given that \( y = 1 \), we can set up the equation:
\[ 1 = \log_2(\log_6(2^{\sqrt{2x+1}} + 4)) \]

### Step 2: Convert the logarithmic equation to exponential form
Using the property of logarithms, we can rewrite the equation in exponential form:
\[ \log_6(2^{\sqrt{2x+1}} + 4) = 2^1 = 2 \]

### Step 3: Convert the second logarithmic equation to exponential form
Now, we can rewrite this equation again in exponential form:
\[ 2^{\sqrt{2x+1}} + 4 = 6^2 \]
Calculating \( 6^2 \):
\[ 2^{\sqrt{2x+1}} + 4 = 36 \]

### Step 4: Solve for \( 2^{\sqrt{2x+1}} \)
Subtract 4 from both sides:
\[ 2^{\sqrt{2x+1}} = 36 - 4 = 32 \]

### Step 5: Convert to base 2
We know that \( 32 = 2^5 \), so we can write:
\[ 2^{\sqrt{2x+1}} = 2^5 \]

### Step 6: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ \sqrt{2x+1} = 5 \]

### Step 7: Square both sides
Squaring both sides gives us:
\[ 2x + 1 = 25 \]

### Step 8: Solve for \( x \)
Subtract 1 from both sides:
\[ 2x = 25 - 1 = 24 \]
Now divide by 2:
\[ x = \frac{24}{2} = 12 \]

### Conclusion
The point on the graph where the y-coordinate is 1 is:
\[ (12, 1) \]